Are arbitrary nonempty intersections of principal filters principal? Suppose $\langle L,\leq\rangle$ is a lattice with join $\sqcup$. Let $F_1$ and $F_2$ be principal filters on $L$. Thus, for $i\in I=\{1,2\}$ there are $x_i\in L$ so that $F_i=\{y\in L:x_i\leq y\}$.
In this situation, $F_1\cap F_2$ is also a principal filter, because $F_1\cap F_2=\{y\in L:x_1\sqcup x_2\leq y\}$. This isn't too hard to see: clearly if $x_1\sqcup x_2\leq y$, then $x_1\leq y$ (so $y\in F_1$) and $x_2\leq y$ (so $y\in F_2$). Thus $\{y\in L: x\sqcup y\leq y\}\subseteq F_1\cap F_2$. On the other hand, if $y\in F_1\cap F_2$, then $x_1\leq y$ (because $y\in F_1$) and $x_2\leq y$ (because $y\in F_2$) and thus $x_1\sqcup x_2\leq y$. So $F_1\cap F_2\subseteq\{y\in L:x_1\sqcup x_2\leq y\}$.
Inductively, if $I$ is finite then $\bigcap_{i\in I} F_i$ is a principal filter. But what if $I$ is arbitrary? In general, of course, the intersection of an arbitrary family of principal filters might be empty (example: take the divisibility poset on $\mathbb{N}$, and let $[p]$ be the principal filter generated by a prime $p$. The intersection of all such filter is clearly empty, since no number is divisible by every prime). Ok, sure. But (now the actual question):

provided each $F_i$ is a principal filter and $\bigcap_{i\in I} F_i$ is nonempty, is it a principal filter?

In all the examples I can think of, ensuring the big intersection is nonempty essentially guarantees that the result is a principal filter. So e.g. in the lattice of subsets of $\mathbb{N}$, let $E$ be the set of all subsets of the even numbers, and let $F_E$ be the set containing, for each $e\in E$, the principal filter generated by $e$. Then, lo and behold, the intersection of all members of $F_E$ is the filter generated by $E$. And clearly if $L$ is complete, then the answer is an easy 'yes'. So I guess I'm really interested in what we can say generally, for any old lattice $L$.
 A: The intersections of principal filters along with the intersections of principal ideals both correspond to the elements of a Dedekind-Macneille completion of a poset $X$.
Let $X$ be a poset. Let $\uparrow,\downarrow:P(X)\rightarrow P(X)$ be the sets where $\uparrow R$ is the collection of all upper bounds of $R$ and $\downarrow R$ is the collection of all lower bounds of $R$.
Let $C=\{\downarrow R\mid R\subseteq X\},D=\{\uparrow R\mid R\subseteq X\}$. In other words, $C$ is the collection of all intersections of principal ideals, and $D$ is the collection of all intersections of principal filters. Then by standard facts about Galois connections, the sets $C,D$ are in one-to-one correspondences with each other by the inverse mappings $C\rightarrow D,R\mapsto\uparrow R$ and $D\rightarrow C,S\mapsto\downarrow S$.
The set $C$ is just the Dedekind-Macneille completion of $X$, and the mapping
$X\mapsto C,x\mapsto\downarrow\{x\}$ is the embedding of $X$ into its Dedekind-MacNeille completion.
We observe that the poset $X$ is complete lattice if the mapping $X\mapsto C,x\mapsto\downarrow\{x\}$ is a bijection, but this happens precisely when the intersection of principal ideals is a principal ideal. Similarly, the poset $X$ is a complete lattice if and only if the intersection of principal filters is a principal filter.
We also observe that $\bigvee_{i\in I}x_i=x$ if and only if $\bigcap_{i\in I}\uparrow x_i=\uparrow x$. By duality, $\bigwedge_{i\in I}x_i=x$ if and only if $\bigcap_{i\in I}\downarrow x_i=\downarrow x.$ Therefore, to look for intersections of principal filters that are non-principal, one should simply look for subsets without any least upper bound. There are plenty of examples of subsets of bounded lattices with bounded subsets that have no least upper bounds including $\mathbb{Q}$ and $\mathbb{R}\setminus\{0\}.$ For a more interesting example, if $X$ is an ultraproduct of structures by a non $\sigma$-complete ultrafilter, then $X$ will be $\aleph_1$-saturated. If $X$ is a $\aleph_1$-saturated poset and $(x_n)_{n\in\omega}$ is a strictly increasing sequence in $X$, then $X$ has no least upper bound.
A: Nope. For a silly counterexample consider the lattice of nonzero real numbers with the usual order. Consider the principal filters $F_i = \left\{y \ne 0: -\frac{1}{i} \le y  \right\}$. Then $F=\bigcap_i F_i $ should be the principal filter generated by zero. But since zero is not in the lattice $F$ is the filter of positive elements. This is nonprincipal.
A: Let $L$ be the lattice of finite and cofinite subsets of $\mathbb N$. Let $\mathcal F_n$ be the principal filter of elements of $L$ that contain the element $n$ (i.e., $\mathcal F_n$ contains the elements of $L$ which dominate the singleton $\{n\}$).
Intersect all $\mathcal F_n$ where $n$ is even. The result will be the filter $\mathcal F$ of the cofinite subsets of $\mathbb N$ that contain all the even numbers. This filter is not principal.
A: There are two good answers already; but I’ll add a little motivation for how one might find the way to them.
If $L$ is complete, then as you say, it’s easy to see that any intersection of principal filters is principal.  Specifically, writing $(a)$ for the principal filter $\{ x \in L \mid x \geq a \}$, we have
$$ \bigcap_{i} (a_i) = (\bigvee_i a_i). $$
But actually, more holds: this formula characterises the join.  For family in an arbitrary lattice, $\bigcap_i (a_i) = (a)$ if and only if $a$ is a supremum for the family of $a_i$’s (just by pushing around definitions).
So finding an intersection of principal filters that’s not principal is precisely equivalent to finding a lattice with a failure of completeness — and following this idea can lead one to Keith’s counterexample.
On the other hand, this formula suggests that intersections of principal filters are “joins that want to exist”, or “formal joins in $L$” — and this idea leads to the fact that such filters form a formal completion of $L$, as described in Joseph van Name’s answer.
